KCET · Maths · Inverse Trigonometric Functions
If
\(\sin ^{-1}\left(\frac{2 a}{1+a^2}\right)+\cos ^{-1}\left(\frac{1-a^2}{1+a^2}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\)
where \(a, x \in(0,1)\), then the value of \(x\) is
- A \(\frac{a}{2}\)
- B \(\frac{2 a}{1+a^2}\)
- C \(\frac{2 a}{1-a^2}\)
- D \(0\)
Answer & Solution
Correct Answer
(C) \(\frac{2 a}{1-a^2}\)
Step-by-step Solution
Detailed explanation
Here,
\(\sin ^{-1}\left(\frac{2 a}{1+a^2}\right)+\cos ^{-1}\left(\frac{1-a^2}{1+a^2}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\)
We know that
\(2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)
and \(2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\)
\(\because 2 \tan ^{-1} a+2 \tan ^{-1} a=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\)
\(2\left[\tan ^{-1}\left(\frac{a+a}{1-a \times a}\right)\right]=2 \tan ^{-1} x\)
\(\Rightarrow \quad \frac{2 a}{1-a^2}=x\)
\(\sin ^{-1}\left(\frac{2 a}{1+a^2}\right)+\cos ^{-1}\left(\frac{1-a^2}{1+a^2}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\)
We know that
\(2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)
and \(2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\)
\(\because 2 \tan ^{-1} a+2 \tan ^{-1} a=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\)
\(2\left[\tan ^{-1}\left(\frac{a+a}{1-a \times a}\right)\right]=2 \tan ^{-1} x\)
\(\Rightarrow \quad \frac{2 a}{1-a^2}=x\)
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